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AP Physics 1 : Motion in One Dimension

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Example Questions

Example Question #31 : Motion In One Dimension

A car is traveling on the freeway at $36 \frac{m}{s}$ when the driver sees traffic stopped 100 m ahead. What acceleration is needed to stop the car safely?

Possible Answers:

$13.0 \frac{m}{s^2}$

$6.5 \frac{m}{s^2}$

$0.18 \frac{m}{s^2}$

$-0.18 \frac{m}{s^2}$

$-6.5 \frac{m}{s^2}$

Correct answer:

$-6.5 \frac{m}{s^2}$

Explanation:

We begin by noting we have been given no information about the time in which the stopping is to occur, only velocities and distance. This points us to the following kinematic equation:

$V_f^2=V_i^2+2a\Delta x$

By using the given information, and noting that coming to a stop implies a final velocity of $0 \frac{m}{s}$ , we can directly substitute the numbers in and solve. It makes sense the final value is negative because the car is being accelerated in the negative direction in order to stop it.

$a=\frac{-(36\frac{m}{s})^2}{2 (100m)}=-6.5 \frac{m}{s^2}$

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Example Question #31 : Motion In One Dimension

An airplane is flying at $37\frac{m}{s}$ . It encounters a tailwind traveling at $9\frac{m}{s}$ . At what speed is the airplane traveling?

Possible Answers:

$37\frac{m}{s}$

$9\frac{m}{s}$

$46\frac{m}{s}$

$28\frac{m}{s}$

Correct answer:

$46\frac{m}{s}$

Explanation:

A tailwind is a wind traveling in the same direction as an object. The find the new speed of the airplane, we add the speed of the tailwind to the speed of the airplane.

$\vec{v}= (37+9 )\frac{m}{s}=46\frac{m}{s}$

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Example Question #32 : Motion In One Dimension

A ball is dropped at time t=0 from 20 meters above the ground.

$g=10\frac{m}{s^2}$

How long does it take for the ball to hit the ground?

Possible Answers:

-2s

Correct answer:

Explanation:

Using the following kinematic equation:

$x=x_{0}+v_{0}t+\frac{1}{2}at^{2}$

is the final position, in meters, ( for ground).

$x_{0}$ is the starting position ( 20m above ground)

$v_{0}$ is the starting velocity $(0\frac{m}{s})$ because the ball is dropped and therefor starts from rest).

is the acceleration due to gravity given by the problem statement.

Input these values into the first ruling equation and note that in this notation up is the positive direction so the acceleration due to gravity is negative:

$0m=20m+(0\frac{m}{s})(t)+\frac{1}{2}(-10\frac{m}{s^{2}})t^{2}$

Then we solve for .

$0=20+0-5t^{2}$

$-20=-5t^{2}$

$4=t^{2}$

t=2s

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Example Question #31 : Motion In One Dimension

A hot air balloon is rising at $2 \frac{m}{s}$ when a bag falls from the edge of the basket.

How far has the bag fallen from where it left the basket after 3 seconds?

$g=10\frac{m}{s^{2}}$

Possible Answers:

49m

30m

39m

45m

26m

Correct answer:

39m

Explanation:

We use the equation:

$x=x_{0}+v_{0}t+\frac{1}{2}at^{2}$

With $x_{0}=0m$ , $v_{0}=2\frac{m}{s}$ , $a=-10\frac{m}{s^{2}}$ , and t=3s we have:

$x=0+3*2+\frac{1}{2}*-10*3^{2}$

x=-39m

The bag has fallen 39 meters from its original position. Note that the question does not ask for the distance the bag is from the basket after 3 seconds so we do not have to take in to account how far the balloon has moved in the 3 seconds. Also, note that the question asks "how far" the bag is from the basket, so we do not have to worry about sign in the answer.

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Example Question #31 : Motion In One Dimension

A rocket ship moves at $200\frac{m}{s}$ when a part breaks off and falls straight down. The rocket continues to move at the same velocity after the part breaks off.

How far below the ship is the part after 5 seconds?

$g=10\frac{m}{s^{2}}$

Possible Answers:

1000m

125m

875m

75m

275m

Correct answer:

125m

Explanation:

We use the equation:

$x=x_{0}+v_{0}t+\frac{1}{2}at^{2}$

With $x_{0}=0m$ , $v_{0}=200\frac{m}{s}$ , $a=-10\frac{m}{s^{2}}$ , and t=5s we have:

$x=0+200*5+\frac{1}{2}*-10*5^{2}$

x=875m

The part moves 875 meters upward in the time it breaks from the ship. Then, we use:

x=vt with $v=200\frac{m}{s}$ and t=5s to find that the ship has moved 1000 meters during the same 5 second period.

So the ship moved upward 1000 meters in the 5 second time period while the part moved 875 meters upward so to find the distance between the ship and the part as the problem statement requested we subtract:

1000m-875m=125m

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Example Question #33 : Motion In One Dimension

A car moving to the right at $20\frac{m}{s}$ accelerates to the right at $3\frac{m}{s^{2}}$ for 14 meters.

What is the cars velocity after the acceleration period?

Possible Answers:

$22\frac{m}{s}$

$29\frac{m}{s}$

$20\frac{m}{s}$

$18\frac{m}{s}$

$26\frac{m}{s}$

Correct answer:

$22\frac{m}{s}$

Explanation:

We use the equation:

$v_{f}^{2}=v_{0}^{2}+2ax$

Where the initial velocity is $v_{o}=20\frac{m}{s}$ , the acceleration is $a=3\frac{m}{s^{2}}$ , and the displacement during the acceleration x=14m .

Plug in those values into our equation and solve.

$v_f^2=(20\frac{m}{s})^2+2*(3\frac{m}{s^2})*14m$

$v_f=\sqrt{400+84}=\sqrt{484}=22\frac{m}{s}$

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Example Question #34 : Motion In One Dimension

Calculate the acceleration, of a race car that can go a distance of 1320ft in 3.5s , assuming the acceleration is constant and the car starts from rest.

$1ft\approx 0.3048m$

Possible Answers:

$215.5 \frac{m}{s^2}$

$5.4\frac{m}{s^2}$

$229.7\frac{m}{s^2}$

$65.7 \frac{m}{s^2}$

Correct answer:

$65.7 \frac{m}{s^2}$

Explanation:

However unrealistic, if the acceleration of the race car were constant, we could calculate a value. We need everything in SI units, so we have to convert the distance to

$1320ft\left(\frac{0.3048m}{1ft}\right )=402.3m$

Using a kinematic equation, we can directly calculate the acceleration ,

$\Delta x = \frac{1}{2} a t^2 \Rightarrow a = \frac{2 \Delta x}{t^2}=\frac{2(402.3m)}{(3.5s)^2}=65.7\frac{m}{s^2}$

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Example Question #11 : Ap Physics B

A block is placed at a height of 18m on a ramp at an incline of 24^o . It slides down from rest with an acceleration of $2.2\frac{m}{s^2}$ due to a frictional force. What is the velocity of the block when it reaches the bottom of incline?

24_deg._incline_block

Possible Answers:

$v_f=8.9\frac{m}{s}$

$v_f=5.1\frac{m}{s}$

$v_f=16\frac{m}{s}$

$v_f=14\frac{m}{s}$

$v_f=12\frac{m}{s}$

Correct answer:

$v_f=14\frac{m}{s}$

Explanation:

To find the final velocity we use the following kinematic equation.

$v_f^2=v_o^2+2a\Delta x$

The block starts from rest, so v_o=0 . The incline is eighteen meters high, so we can find the distance (hypotenuse) as follows:

$\Delta x=\frac{18m}{sin24^\circ}$

$\Delta x = 44.25m$

We know that the acceleration is $a=2.2\frac{m}{s^2}$ .

Plug in the values to find the velocity at the bottom of the incline.

$v_f^2=(0\frac{m}{s})^2+2(2.2\frac{m}{s^2})(44.25m)$

$v_f=\sqrt{194.7\frac{m^2}{s^2}}=14\frac{m}{s}$

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Example Question #37 : Motion In One Dimension

Object A is traveling at $<10,0>\frac{m}{s}$ . It is hit from behind by object B traveling at $<15,0>\frac{m}{s}$ . If they both have a mass of 25kg , determine their final velocity if they stick together.

Possible Answers:

$<0,12.5>\frac{m}{s}$

$<15.0,0>\frac{m}{s}$

$<15,15>\frac{m}{s}$

$<0,0>\frac{m}{s}$

$<12.5,0>\frac{m}{s}$

Correct answer:

$<12.5,0>\frac{m}{s}$

Explanation:

$\overrightarrow{P}_{Total}=\overrightarrow{P}_1+\overrightarrow{P}_2...$

Definition of momentum:

$\overrightarrow{P}=m\overrightarrow{v}$

Combine equations:

$\overrightarrow{P}_{Total}=m_1\overrightarrow{v}_1+m_2\overrightarrow{v}_2$

Plug in values:

$\overrightarrow{P}_{Total}=25*<10,0>+25*<15,0>$

$\overrightarrow{P}_{Total}=<250,0>+<375,0>$

$\frac{\overrightarrow{P}_{Total}}{m}=\frac{<625,0>kg\frac{m}{s}}{50kg}$

$\overrightarrow{v}=<12.5,0>\frac{m}{s}$

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Example Question #311 : Ap Physics 1

A ball is dropped from a building of height 45m. How long does it take for the ball to hit the ground? Ignore air resistance.

Possible Answers:

10s

Correct answer:

Explanation:

Let's first list the values we know.

$a=10\frac{m}{s^2}$

$v_o=0\frac{m}{s}$

d=45m

We are looking for the time it takes for this ball to reach the ground, thus we are looking for . The kinematics equation we need to use is:

$d=v_ot +\frac{1}{2}at^2$

Since the initial velocity is zero, the v_ot term disappears. Thus we are left with:

$45m=\frac{1}{2}*10\frac{m}{s^2}t^2$

Solve for time.

t=3s

We only take the positive value in our calculation since time cannot be negative.

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AP Physics 1 : Motion in One Dimension

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #31 : Motion In One Dimension

Example Question #31 : Motion In One Dimension

Example Question #32 : Motion In One Dimension

Example Question #31 : Motion In One Dimension

Example Question #31 : Motion In One Dimension

Example Question #33 : Motion In One Dimension

Example Question #34 : Motion In One Dimension

Example Question #11 : Ap Physics B

Example Question #37 : Motion In One Dimension

Example Question #311 : Ap Physics 1

All AP Physics 1 Resources

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